3.1 Periodicity Flashcards
Chromium and copper exception
- ₂₄Chromium: acutal: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵ or [Ar] 3d54s1, expected [Ar] 4s2 3d4
₂₉Copper: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰ or [Ar] 3d104s1, expeted [Ar] 4s2 3d9
Because having one electron in each of the five d orbitals is more stable than having 4 filled and 1 empty/partially empty
a)state the group of alkali metal
b)state group of alkaline metal
c)state group of s,p,d,f
4)state the group of noble gas
states actinides and actinides
group
columns and number of valence electrons
Lanthanide
F block and part of 6th peroid
period
rows and shows number of occupied energy level or the number of outermost occupied level
Actinides
F block and part of 7th peroid
Atomic radius
The atomic radius is ½ the distance between two adjacent identical atoms in a covalent bond
Ionic radius
Distance between the nucleus and the outermost electrons
electron affinity
The energy change to form a mole of gaseous anions
- 𝑋(𝑔) + 𝑒 − → 𝑋(𝑔)
- 𝑋-1(𝑔) + 𝑒- − → 𝑋-2(𝑔)
- 𝑋-2(𝑔) + 𝑒- − → 𝑋-3(𝑔)
metalloids
An element that possesses some of the properties of a metal and some of a non-metal. While metal oxides tend to be basic and non-metal oxides tend to be acidic, metalloid oxides such as aluminum oxide can be amphoteric.
Define Periodic table
Periodicity refers to repeating trends or patterns of physical and chemical properties in elemen
State three properties that are characteristic of transitional elements
- variable oxidation state
- form complex ions with ligands
- form coloured compounds
- may act as catalyst (as a element or compound)
- Paramagnetic, diamagnetic…
Describe the trend atomic radii across the period
- number of protons increases (nuclear energy)
- number of electron shell/sheilding effect is kept the same
- effective nuclear energy increases and increased electrostatic attraction and distance between postively charged nuclues and valence electrons decreases
- smaller atomic radii
Decribe the atomic radii down the group
- increased energy level/shielding
- decreased electrostatic attraction and distance between positively chaged nucleus and valence electrons
- increased atomic radii
Describe the ionic radii down the group
No pattern due to varuable oxidation state of transitional metal
Why is Mg2+ smaller than Na+
- they are isoelectronic (same number of electron)
- Mg2+ has smalller nuclear charge (#protons)
Why is Cl- bigger than Cl?
- Anions are bigger than parent atom
- addiction of electron causing increased e-e reputsion causing the electrons to move further apart
- increased radii
Why is Mg2+ smalller than Mg?
- Cations are bigger than parent atom
- group 14~17: cations
- losses an outer shell
- decreased radii
Describe the ionic radius across period three
- Na+, Mg2+, Al3+, Si4+ decrease in ionic radius due to increased effective nuclear charge
- Si4-, big increase in ionic radius due to added energy level and more electrons than protons , decreased electroststic attraction
*Si4-, P3-, S2-, Cl- , decrease ionic radius due to increased effective nuclear charge
Explain ionization trend of Ca and Ti
Ca: [Ar] 4s²→ big jump after removing second e due to transition form fourth to third energy level
Ti: [Ar] 3d² 4s² → I.E increases gradually (4s and 3d similar) and does not jump till 5th electron, which explain varible oxidation state of Ti being 2+, 3+, 4+ but not +5
State the 2 exceptions to the ionization energy trend across the period
- Drop→ higher in energy level, further away from the nucleus
- Rise→ increased number of protons, effective nuclear charge
- Be~B→ The outer electron in B is in a 2P sublevel and therefore has higher energy and further away from the nucleus than the outer electron in Be which is in 2S usblevel. Therefore the 2P sublevel requires less enrgy to remove than than from 2S
- N~O→ three 2P electrons in nitrogens are unpaired. 4 electrons in 2P with one doubly filled in Oxygen experiences e-e repulsion therefore eaiser to remove
Explain the trend of succesive ionization energy of Na
- 1s2 2s2 2p6 3s1
- 3s1→ higher in energy level, further away from the nucleus → easiest to remove
- 3s1~2p6 → one energy level lost/lower in energy level→ less sheilding by inner electrons and closer to the nucleus
- 2s1~2p6 → number of protons stays same , more attraction between remaining electrons → higher effective nuclear chage → less e-e repulsion
First ionization energy
The minimum amount of energy needed to remove one moles of electrons from one moles of an element in the gaseous state
Always endothermic (energy absorbed) 𝑋(𝑔) → 𝑋(𝑔)+ + 𝑒−
- the metal have low I.E. and non metal have high I.E
trend for succesive ionization energy at same energy level of an atom
There is an increase in successive ionization energies. more electrons are removed the remaining electrons
experience an increasing effective nuclear charge and are held closer to the nucleus and hence more tightly
trend for large jumps in succesive ionization of an atom
a change to a new inner shell, closer to the nucleus/lower in energy level/with the electrons held more strongly
State reason to variables oxidation state of Al and Cr
*aluminium +3 +1and +2, electrons removed frm [Ne] 3s² 3p1 to make bondings with more stable atom
* Cr +1, +2, +3, +4, +5, +6, electron removed from [Ar]4s13d
Trend for electron affinity
- Group 17→ highest electron affinity → big incomplete energy level with highest effective nuclear charge
- Group 1 → lowest electron affinity → effetive nuclear charge of 1+
- metals have low affinity and non metal have high affinity
- Electronegativity
- Explain the polarity of S-O bond
- abillity of itos atom to attract electrons in covalent bond
- oxygen hav ehigher electron density/negative partial dipole
- Sulfur have low electron density/positive paritial dipole